Cambridge A Level Mathematics 9709 — 2005 Oct/Nov Paper 5 · Variant 1

1 A uniform solid cone has vertical height 28 cm and base radius 6 cm. The cone is held with a point of the circumference of its base in contact with a horizontal table, and with the base making an angle of θ◦with the horizontal (see diagram). When the cone is released, it moves towards the equilibrium position in which its base is in contact with the table. Show that θ < 40.6, correct to 1 decimal place. [3]

Mark scheme: 1 (28/4) 6 B1 x = ¼ 28 (θ + tan-1(7/6) < 90) M1 For using θ + tan −1 ( x / 6) < 90 θ < 40.6 A1 3

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2 An aircraft flies horizontally at a constant speed of 220 m s−1. Initially it is flying due east. On reaching a point A it flies in a circular arc from A to B, taking 50 s. At B the aircraft is flying due south (see diagram). (i) Show that the radius of the arc is approximately 7000 m. [3] (ii) Find the magnitude of the acceleration of the aircraft while it is flying between A and B. [2]

Mark scheme: 2 (i) (½ π = ω 50 or L = 220×50) M1 For using θ = ω t or L = vt [220 = ( π / 100 )r or M1 For using v = ω r or L = rθ 11000 = r( ½π )] Radius is approx. 7000 m A1 3 (ii) M1 For using a = v2/r or a = ω 2r Acceleration is 6.91ms-2 A1 2

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3 A uniform lamina ABCD is in the form of a trapezium in which AB and DC are parallel and have lengths 2 m and 3 m respectively. BD is perpendicular to the parallel sides and has length 1 m (see diagram). (i) Find the distance of the centre of mass of the lamina from BD. [3] The lamina has weight W N and is in equilibrium, suspended by a vertical string attached to the lamina at B. The lamina rests on a vertical support at C. The lamina is in a vertical plane with AB and DC horizontal. (ii) Find, in terms of W, the tension in the string and the magnitude of the force exerted on the lamina at C. [3]

Mark scheme: 3 (i) M1 For obtaining an equation in x by taking moments about, for example, BD 0.6W×1 – 0.4W×(2/3) = W x or A1 Any correct equation in x , with or ½( 3×1) ×1 – ½ (2×1) ×(2/3) = without W throughout. (3/2 + 1) x Distance is 1/3 m A1 3 (ii) 3T = (8/3)W or 3FC = (1/3)W M1 For taking moments about C or about BD Tension is 8W/9 or force at C = W/9 A1 ft ft for T = (1 - x /3)W or Force at C = W/9 or tension is 8W/9 A1 ft 3 FC = ( x /3)W GCE A/AS LEVEL – November 2005 9709, 8719 5

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4 A particle is projected from horizontal ground with speed u m s−1 at an angle of θ◦above the horizontal. The greatest height reached by the particle is 10 m and the particle hits the ground at a distance of 40 m from the point of projection. In either order, (i) find the values of u and θ, (ii) find the equation of the trajectory, in the form y = ax −bx2, where x m and y m are the horizontal and vertical displacements of the particle from the point of projection. [7]

Mark scheme: 4 (i) u2sin2 θ ÷ 2g = 10 or B1 Using maximum height ½ (usinθ + 0)T = 10 2u2sinθ cosθ ÷ g = 40 or B1 Using range (or half range) u(2T)cosθ = 40 or uTcosθ = 20 2 M1 For eliminating u2 or uT sin θ 2 g (10) [ = sin θ cos θ [ g ( 40) ÷ 2] sin θ 2 × 10 or = ] cos θ 40 ÷ 2 θ = 45 A1 u2 = 20×10÷ ½ u = 20 or A1 5 u ÷ 2 = gT and uT ÷ 2 2 = 10 u = 20 (ii) y = xtan45o – gx2÷ (2×202cos245o) M1 For substituting for u and θ in the general equation y = x – x2/40 A1 2 OR 4 (ii) y = kx(40 – x) M1 For quadratic equation with roots x = 0 and x = 40 10 = 400k M1 For using y = 10 when x = 20 y = x – x2/40 A1 3 (i) 1 = tanθ M1 For equating coefficients of x with that of general form θ =45o A1 1 10 M1 For equating coefficients of x2 with − = − 2 40 2 u × 1 / 2 that of general form with θ =45o substituted u = 20 A1 4

More questions on Kinematics of motion in a straight line

5 A particle P of mass 0.2 kg is attached to the mid-point of a light elastic string of natural length 5.5 m and modulus of elasticity λ N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 6 m apart. P is held at rest at a point 1.25 m vertically above the mid-point of AB and then released. P travels a distance 5.25 m downwards before coming to instantaneous rest (see diagram). By considering the changes in gravitational potential energy and elastic potential energy as P travels downwards, find the value of λ. [8]

Mark scheme: 5 Loss in GPE = 0.2g×5.25 (10.5 J) B1 AP is 3.25 initially and 5 finally B1 M1 For using EE = λ x2 ÷ (2L) L must be correct (2.75 or 5.5) For any correct expression for Initial EPE or for Final EPE A1 ft ft incorrect AP [2×0.52λ ÷ (2×2.75) for initial or 2×2.252λ ÷ (2×2.75) for final] Gain in EPE = (81-4)λ /44 =1.75λ A1 Any correct expression M1 For applying the principle of conservation of energy 1.75λ = 10.5 A1 ft For any correct equation in λ , ft only if initial and final EPE are used λ = 6 A1 8 GCE A/AS LEVEL – November 2005 9709, 8719 5

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6 A horizontal circular disc of radius 4 m is free to rotate about a vertical axis through its centre O. One end of a light inextensible rope of length 5 m is attached to a point A of the circumference of the disc, and an object P of mass 24 kg is attached to the other end of the rope. When the disc rotates with constant angular speed ω rad s−1, the rope makes an angle of θ radians with the vertical and the tension in the rope is T N (see diagram). You may assume that the rope is always in the same vertical plane as the radius OA of the disc. (i) Given that cos θ = 2425, find the value of ω. [5] (ii) Given instead that the speed of P is twice the speed of the point A, find (a) the value of T, [3] (b) the speed of P. [2]

Mark scheme: 6 (i) Radius of path = 4 + 5×7/25 B1 ( =5.4m) (T×(24/25) = 24×10) (T = 250) M1 For resolving forces vertically M1 For applying Newton’s second law horizontally and using a = ω 2r 24ω 2 ×5.4 = 250×(7/25) A1 ft ω = 0.735 A1 5 (ii)(a) Radius of path = 2×4 B1 Using v is proportional to r sinθ = 0.8 B1 T = 400 B1ft 3 ft wrong θ (b) 24 v 2 4 M1 For applying Newton’s second law ( = 400 × ) 8 5 horizontally and using a = v2/r Speed is 10.3 ms-1 A1 2

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7 A particle of mass 0.25 kg moves in a straight line on a smooth horizontal surface. A variable resisting force acts on the particle. At time t s the displacement of the particle from a point on the line is x m, and its velocity is (8 −2x) m s−1. It is given that x = 0 when t = 0. (i) Find the acceleration of the particle in terms of x, and hence find the magnitude of the resisting force when x = 1. [3] (ii) Find an expression for x in terms of t. [6] (iii) Show that the particle is always less than 4 m from its initial position. [2]

Mark scheme: 7 (i) a = (8 – 2x)(-2) = -16 +4x B1 Any correct form -R = 0.25(-16 + 4×1) M1 For using Newton’s second law and substituting for x Magnitude of the force is 3 N A1 3 (ii) dx B1 ∫ dt = ∫ 8 − 2 x M1* For attempting to integrate t = -1/2 ln(8 – 2x) (+C) A1 (C = 1/2 ln8) M1*dep For using x = 0 when t = 0 to find C 8 2 t 8 2 t = ln ⇒ e = 8 − 2 x 8 − 2 x M1*dep For converting to exponential form x = 4(1 – e-2t) A1 6 (iii) t ≥ 0 M1 0 < e −t2 ≤ 1 0 ≤ 1 − e −t2 < 1 0 ≤x < 4 A1 2

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